ID 23

some subgroup H C G, and although there are usually many subgroups of

G whose image in G is the given subgroup if, exactly one of them contains

N. If H C G is arbitrary, we see that HN = H N — H since overbar is

a homomorphism and iV is its kernel. It follows that HN is the unique

subgroup containing N whose image in G is H. In particular, since indices

of corresponding subgroups are equal, we have \G : H\ — \G : NH\ for all

subgroups H C G.

The correspondence theorem also yields information about normality.

If T V C H C K C G, then if K if and only if H ~K. In particular,

if N C if, then since if C Nc(if), we see that if Nc(if) and we have

Ns(if) Q N^(il). In fact, equality holds here. To see this, observe that

since N^(ff) is a subgroup of G, it can be written in the form U for some

(unique) subgroup U with N C [7 C G. Then H U, and so H U and

[7 C N

G

(if). This yields I%(ff) = C f C N

G

(ff), as claimed.

We will use the bar convention in the following proof.

Proof of Theorem 1.22. Since G is nilpotent, it has (by definition) a cen-

tral series {Ni | 0 i r}, and we have N0 = 1 C if and Nr = G £ H. It

follows that there is some subscript k with 0 k r such that Nk C H but

Nk+i % H. We will show that in fact, Nk+t C N

G

(if), and it will follow

that NG(H) if, as required.

Write G = G/Nk and use the bar convention. Since the subgroups Ni

form a central series, we have

N^~i C Z(G) C Ny(F) = N ^ ) ,

where the equality holds because Nk C if. Now because iV^ C No(if),

we can remove the overbars to obtain A^+i C Nc(if). The proof is now

complete. •

We return now to p-groups, with another application of Theorem 1.19.

1.23. Lemma. Let P be a finite p-group and suppose that N M are

normal subgroups of P. Then there exists a subgroup L P such that N C

LCM and \L : N\ =p.

Proof. Write P = P/N and note that M is nontrivial and normal in P.

Now Z(P) flM is nontrivial by Theorem 1.19, and so this subgroup contains

an element of order p. (Choose any nonidentity element and take an appro-

priate power.) Because our element is central and of order p, it generates a

normal subgroup of order p, and we can write L to denote this subgroup,

where N C L. Now LCM, and thus N C L C M, as wanted. Also, as

L P, we see that L P. Finally, \L : N\ = \L\ = p, as required. •